Two buses starting from two different places A and B simultaneously, travel towards each other and meet after a specified time. If the bus starting from A is delayed by 20 minutes, they meet 12 minutes later than the usual time. If the speed of the bus starting from A is 60km/hr, what is the speed of the second bus?
My attempt:
Let t be the usual time of meeting and s(A) and s(B) be the speeds of Bus A and Bus B respectively, then
Applying relative velocity concept (stopping bus B i.e. taking it's speed =0 and adding it to speed of A), i got:
s(A)= 60 (given)
(60 + s(B))t = d (distance between the buses)
Now if bus A is delayed by 20 mins, then B's travelling time is 20 mins more than A's to cover the same distance.
t(B) = t(A) + 20/60
hence, t(B)= t(A) + 1/5
distance covered by A = t(A) * 60
distance covered by B = t(B) * s(B) = (t(A) +1/5)*s(B)
The problem: How do i relate the delay of 12 mins in their meeting time?
Best Answer
When the buses meet, Bus A has travelled $8$ fewer minutes than usual, and therefore $8$ fewer km.
These $8$ km were travelled by B in the $12$ extra minutes. So the speed of B is $\dfrac{8}{12}$ km per minute.